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11.14-2. Some Special Cases

               Assume that
                                                         n

                                         K(x, t)= k(x, t) –  g k (x)h k (t),                (6)
                                                         k=1
               i.e., in formula (3) we have a ij = 0 for i ≠ j and a ii = 1. For this case, the resolvent is equal to


                               r(x, t; λ) ϕ 1 (x) ··· ϕ n (x)      1+ λb 11  λb 12  ···  λb 1n

                           1   ψ 1 (t)  1 + λb 11 ···  λb 1n           λb 21  1+ λb 22 ···  λb 2n
                R(x, t; λ)=      .       .    .     .      , ∆ ∗ =     .   .   .     .     . (7)
                                 .       .     .    .              .       .     .   .
                                 .       .    .     .              .       .    .    .
                          ∆ ∗
                               ψ n (t)  λb n1  ··· 1+ λb nn       λb n1  λb n2  ··· 1+ λb nn

                   Moreover, assume that k(x, t) = 0, i.e., the kernel K(x, t) is degenerate:
                                                      n

                                            K(x, t)= –   g k (x)h k (t).                    (8)
                                                      k=1
                   In this case it is clear that r(x, t; λ) = 0 and, by virtue of (7),

                                                                   b
                               ϕ k (x)= g k (x),  ψ k (x)= h k (x),  b ij =  g j (x)h i (x) dx.
                                                                 a
               Therefore, the resolvent becomes
                                                 0     g 1 (x)  ···  g n (x)

                                             1   h 1 (t)  1 + λb 11  ···  λb 1n

                                  R(x, t; λ)=     .      .     .      .       .             (9)
                                                  .      .      .     .
                                                  .      .      .     .
                                            ∆ ∗

                                                h n (t)  λb n1  ··· 1+ λb nn
                   Now we consider an integral equation with some kernel Q(x, t). On the interval (a, b)we
               arbitrarily choose points x 1 , ... , x n and t 1 , ... , t n , and in relation (3) we set
                           k(x, t)=0,  g k (x)= Q(x, t k ),  h k (t)= –Q(x k , t),  a ij = Q(x i , t j ).

               In this case it is clear that r(x, t; λ) = 0, and the kernel K(x, t) acquires the form
                                  0

                                       Q(x, t 1 ) ··· Q(x, t n )
                                                                      Q(x 1 , t 1 ) ··· Q(x 1 , t n )
                            1   Q(x 1 , t) Q(x 1 , t 1 ) ··· Q(x 1 , t n )       .  .  .
                   K(x, t)=       .       .    .      .     ,   D =      . .   .  .  . .    .
                            D     . .     . .   . .   . .

                               Q(x n , t) Q(x n , t 1 ) ··· Q(x n , t n )
                                                                     Q(x n , t 1 ) ··· Q(x n , t n )
               It is convenient to rewrite this formula in the form

                                                 Q(x, t)  Q(x, t 1 )  ···  Q(x, t n )
                                             1   Q(x 1 , t)  Q(x 1 , t 1 )  ···  Q(x 1 , t n )

                             K(x, t)= Q(x, t) –     .        .     .       .      .        (10)
                                             D      .        .      . .    .
                                                    .        .             .

                                                Q(x n , t)  Q(x n , t 1 )  ··· Q(x n , t n )
                   The kernel K(x, t) is degenerate and, moreover, it coincides with the kernel Q(x, t)onthe
               straight lines x = x i , t = t j (i, j =1, ... , n). Indeed, if we set x = x i or t = t j , then the determinant in
               the numerator of the second term has two equal rows or columns and hence vanishes, and therefore,
                                      K(x i , t)= Q(x i , t),  K(x, t j )= Q(x, t j ).




                 © 1998 by CRC Press LLC








               © 1998 by CRC Press LLC
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